Dãy trên có 2004x

2004x+(1+2+3+4+...+2003)=2004

2004x+[(2003+1)x{(2003-1):1+1]:2}=2004

2004x+2007006=2004

2004x=2004-2007006

2004x=-2005002

x=-2005002:2004

x=-2001/2

Chúc em học tốt^^

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

\(\Rightarrow x+2004=0\)

\(x=0-2004\)

\(x=-2004\)

<=> 2004x + (2003+1)x2003;2=2004

<=> 2004x = -2005002

<=> x= -1000,5

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=2004\)

\(\Leftrightarrow x\times2003+\left(1+2+3+...+2003\right)=2004\)

\(\Leftrightarrow2003x+\frac{\left(1+2003\right)2003}{2}=2004\)

\(\Leftrightarrow2003\left(x+1002\right)=2004\)

\(x+1002=\frac{2004}{2003}\)

\(x=\frac{2004-2007006}{2003}=\frac{2005002}{2003}\)

x + (x+1) +(x+2) +...+(x+2003)=2004

=> (x+0) + (x+1) +(x+2) +...+(x+2003)=2004

<=> có 2004 cặp

=> (x+x+x+...+x) + (0+1+2+...+2003) = 2004

=> 2004x + 2017026 = 2004

2004x = 2004 - 2017026

2004x = -2015022

x = -2015022 : 2004

x = -1005,5

x+(x+1)+(x+2)+...+(x+2003)=2004

<=>(x+0)+(x+1)+(x+2)+...+(x+2003)=2004

=>Có 2004 cặp số

<=>(x+x+x+...+x)+(0+1+2+...2003)=2004

<=>2004x+2017026=2004

2004x=2004-2017026

2004x=-2015022

x=-2015022:2004

x=-1005,5

Vậy x =-1005,5